A069129 -id:A069129 - cp's OEIS frontend

A139277 a(n) = n(8n+5).

0, 13, 42, 87, 148, 225, 318, 427, 552, 693, 850, 1023, 1212, 1417, 1638, 1875, 2128, 2397, 2682, 2983, 3300, 3633, 3982, 4347, 4728, 5125, 5538, 5967, 6412, 6873, 7350, 7843, 8352, 8877, 9418, 9975, 10548, 11137, 11742, 12363, 13000

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139273 in the same spiral.

Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250.

Cf. A139271, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Table[n (8 n + 5), {n, 0, 50}] (* Bruno Berselli, Aug 22 2018 *)
LinearRecurrence[{3,-3,1},{0,13,42},50] (* Harvey P. Dale, Dec 04 2018 *)

a(n)=n*(8*n+5) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 5*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x*{c+8 + (8-c)*x}/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 3 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
Sum_{n>=1} 1/a(n) = (sqrt(2)-1)*Pi/10 - 4*log(2)/5 + sqrt(2)*log(sqrt(2)+1)/5 + 8/25. - Amiram Eldar, Mar 18 2022
E.g.f.: exp(x)*x*(13 + 8*x). - Elmo R. Oliveira, Dec 15 2024

A195314 Centered 28-gonal numbers.

Original entry on oeis.org

1, 29, 85, 169, 281, 421, 589, 785, 1009, 1261, 1541, 1849, 2185, 2549, 2941, 3361, 3809, 4285, 4789, 5321, 5881, 6469, 7085, 7729, 8401, 9101, 9829, 10585, 11369, 12181, 13021, 13889, 14785, 15709, 16661, 17641, 18649, 19685, 20749, 21841, 22961, 24109, 25285, 26489

Offset: 1

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 1, in the direction 1, 29, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Semi-axis opposite to A144555 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195145.
Cf. A003154, A069129, A069133, A069190, A195315, A195316, A195317, A195318.
Cf. A069127, A118277, A144555.

[(14*n^2-14*n+1): n in [1..50]]; // Vincenzo Librandi, Sep 19 2011

Table[14n^2-14n+1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,29,85},50]

a(n)=14*n^2-14*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 14*n^2 - 14*n + 1.
G.f.: -x*(1 + 26*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 18 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Oct 01 2011
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(5/7)*Pi/2)/(2*sqrt(35)). - Amiram Eldar, Feb 11 2022
From Elmo R. Oliveira, Nov 14 2024: (Start)
E.g.f.: exp(x)*(14*x^2 + 1) - 1.
a(n) = 2*A069127(n) - 1. (End)

A195315 Centered 32-gonal numbers.

Original entry on oeis.org

1, 33, 97, 193, 321, 481, 673, 897, 1153, 1441, 1761, 2113, 2497, 2913, 3361, 3841, 4353, 4897, 5473, 6081, 6721, 7393, 8097, 8833, 9601, 10401, 11233, 12097, 12993, 13921, 14881, 15873, 16897, 17953, 19041, 20161, 21313, 22497, 23713, 24961, 26241, 27553, 28897, 30273

Offset: 1

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 1, in the direction 1, 33, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. Semi-axis opposite to A016802 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195146.
Cf. A003154, A069129, A069133, A069190, A195314, A195316, A195317, A195318.
Cf. A016802, A074377.

[(16*n^2-16*n+1): n in [1..50]]; // Vincenzo Librandi, Sep 19 2011

Table[16*n^2 - 16*n + 1, {n, 1, 41}] (* Amiram Eldar, Feb 11 2022 *)
LinearRecurrence[{3,-3,1},{1,33,97},50] (* Harvey P. Dale, Feb 11 2024 *)

a(n)=16*n^2-16*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 16*n^2 - 16*n + 1.
G.f.: -x*(1 + 30*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 18 2011
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(3)*Pi/4)/(8*sqrt(3)). - Amiram Eldar, Feb 11 2022
From Elmo R. Oliveira, Nov 14 2024: (Start)
E.g.f.: exp(x)*(16*x^2 + 1) - 1.
a(n) = 2*A069129(n) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A195316 Centered 36-gonal numbers.

Original entry on oeis.org

1, 37, 109, 217, 361, 541, 757, 1009, 1297, 1621, 1981, 2377, 2809, 3277, 3781, 4321, 4897, 5509, 6157, 6841, 7561, 8317, 9109, 9937, 10801, 11701, 12637, 13609, 14617, 15661, 16741, 17857, 19009, 20197, 21421, 22681, 23977, 25309, 26677, 28081, 29521, 30997, 32509

Offset: 1

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 1, in the direction 1, 37, ..., in the square spiral whose vertices are the generalized hendecagonal numbers A195160. Semi-axis opposite to A195321 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
John Elias, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195147.
Cf. A003154, A069129, A069133, A069190, A195314, A195315, A195317, A195318.
Cf. A069131, A195160, A195321.

[(18*n^2-18*n+1): n in [1..50]]; // Vincenzo Librandi, Sep 19 2011

Table[18*n^2 - 18*n + 1, {n, 1, 40}] (* Amiram Eldar, Feb 11 2022 *)

a(n)=18*n^2-18*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 18*n^2 - 18*n + 1.
G.f.: -x*(1 + 34*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 18 2011
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(7)*Pi/6)/(6*sqrt(7)). - Amiram Eldar, Feb 11 2022
From Elmo R. Oliveira, Nov 14 2024: (Start)
E.g.f.: exp(x)*(18*x^2 + 1) - 1.
a(n) = 2*A069131(n) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A195318 Centered 44-gonal numbers.

Original entry on oeis.org

1, 45, 133, 265, 441, 661, 925, 1233, 1585, 1981, 2421, 2905, 3433, 4005, 4621, 5281, 5985, 6733, 7525, 8361, 9241, 10165, 11133, 12145, 13201, 14301, 15445, 16633, 17865, 19141, 20461, 21825, 23233, 24685, 26181, 27721, 29305, 30933, 32605, 34321, 36081, 37885, 39733

Offset: 1

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 1, in the direction 1, 45, ..., in the square spiral whose vertices are the generalized tridecagonal numbers A195313. Semi-axis opposite to A195323 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195149.
Cf. A003154, A069129, A069133, A069190, A195314, A195315, A195316, A195317.
Cf. A069173, A195313, A195323.

[22*n^2 - 22*n + 1: n in [1..50]]; // Vincenzo Librandi, Sep 21 2011

Table[22n^2-22n+1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,45,133},50] (* Harvey P. Dale, Mar 16 2019 *)

a(n)=22*n^2-22*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 22*n^2 - 22*n + 1.
Sum_{n>=1} 1/a(n) = Pi*tan(3*Pi/(2*sqrt(11)))/(6*sqrt(11)). - Amiram Eldar, Feb 11 2022
G.f.: -x*(1+42*x+x^2)/(x-1)^3. - R. J. Mathar, May 07 2024
From Elmo R. Oliveira, Nov 15 2024: (Start)
E.g.f.: exp(x)*(22*x^2 + 1) - 1.
a(n) = 2*A069173(n) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.

Original entry on oeis.org

2, 7, 18, 31, 50, 71, 98, 127, 162, 199, 242, 287, 338, 391, 450, 511, 578, 647, 722, 799, 882, 967, 1058, 1151, 1250, 1351, 1458, 1567, 1682, 1799, 1922, 2047, 2178, 2311, 2450, 2591, 2738, 2887, 3042, 3199, 3362, 3527, 3698, 3871, 4050, 4231, 4418, 4607, 4802

Offset: 0

Bruno Berselli, Sep 21 2011 - based on remarks and sequences by Omar E. Pol

Sequence found by reading the numbers in increasing order on the vertical line containing 2 of the square spiral whose vertices are the triangular numbers (A000217) - see Pol's comments in other sequences visible in this numerical spiral.

Also A077591 (without first term) and A157914 interleaved.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms: An origin of A195605.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000217, A077591, A157914.
Cf. A047621 (contains first differences), A016754 (contains the sum of any two consecutive terms).
Cf. A033585, A069129, A077221, A102083, A139098, A139271-A139277, A139592, A139593, A188135, A194268, A194431, A195241 [incomplete list].

[(4*n*(n+2)+(-1)^n+3)/2: n in [0..48]];

CoefficientList[Series[(2 + 3 x + 4 x^2 - x^3) / ((1 + x) (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{2,0,-2,1},{2,7,18,31},50] (* Harvey P. Dale, Jan 21 2017 *)

for(n=0, 48, print1((4*n*(n+2)+(-1)^n+3)/2", "));

G.f.: (2+3*x+4*x^2-x^3)/((1+x)*(1-x)^3).
a(n) = a(-n-2) = 2*a(n-1)-2*a(n-3)+a(n-4).
a(n) = A047524(A000982(n+1)).
Sum_{n>=0} 1/a(n) = 1/2 + Pi^2/16 - cot(Pi/(2*sqrt(2)))*Pi/(4*sqrt(2)). - Amiram Eldar, Mar 06 2023

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

Original entry on oeis.org

0, 15, 1485, 145530, 14260470, 1397380545, 136929032955, 13417647849060, 1314792560174940, 128836253249295075, 12624638025870742425, 1237085690282083462590, 121221773009618308591410, 11878496669252312158495605, 1163971451813716973223977895

Offset: 0

Charlie Marion, Feb 15 2012

For n > 1, a(n) = 98*a(n-1) - a(n-2) + 15. In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n). Then b(0) = 0, b(1) = A014105(m) and for n > 1, b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m). Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).

			2*0 = 3*0.
2*15 = 3*10.
2*1485 = 3*990.
2*145530 = 3*97020.

Colin Barker, Table of n, a(n) for n = 0..500
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008.

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 17, 5, 1, 1, 99, 49, 7, 1, 1, 577, 485, 97, 9, 1, 1, 3363, 4801, 1351, 161, 11, 1, 1, 19601, 47525, 18817, 2889, 241, 13, 1, 1, 114243, 470449, 262087, 51841, 5291, 337, 15, 1, 1, 665857, 4656965, 3650401, 930249, 116161, 8749, 449, 17, 1

Offset: 0

Seiichi Manyama, Dec 26 2018

			Square array begins:
   1,  1,   1,    1,      1,       1,         1, ...
   1,  3,  17,   99,    577,    3363,     19601, ...
   1,  5,  49,  485,   4801,   47525,    470449, ...
   1,  7,  97, 1351,  18817,  262087,   3650401, ...
   1,  9, 161, 2889,  51841,  930249,  16692641, ...
   1, 11, 241, 5291, 116161, 2550251,  55989361, ...
   1, 13, 337, 8749, 227137, 5896813, 153090001, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-3 give A000012, A005408, A069129(n+1), A322830.
Rows 0-9 give A000012, A001541, A001079, A011943(n+1), A023039, A077422, A097308, A068203, A056771, A078986.
Main diagonal gives A173174.
A(n-1,n) gives A173148(n).
Cf. A322699, A322747.

A[0, k_] := 1; A[n_, k_] := Sum[Binomial[2 k, 2 j]*(n + 1)^(k - j)*n^j, {j, 0, k}]; Table[A[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 26 2018 *)

a(n) = 2 * A322699(n) + 1.
A(n,k) + sqrt(A(n,k)^2 - 1) = (sqrt(n+1) + sqrt(n))^(2*k).
A(n,k) - sqrt(A(n,k)^2 - 1) = (sqrt(n+1) - sqrt(n))^(2*k).
A(n,0) = 1, A(n,1) = 2*n+1 and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) for k > 1.
A(n,k) = T_{k}(2*n+1) where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = 2*n+1.

A194268 a(n) = 8n^2 + 7n + 1.

Original entry on oeis.org

1, 16, 47, 94, 157, 236, 331, 442, 569, 712, 871, 1046, 1237, 1444, 1667, 1906, 2161, 2432, 2719, 3022, 3341, 3676, 4027, 4394, 4777, 5176, 5591, 6022, 6469, 6932, 7411, 7906, 8417, 8944, 9487, 10046, 10621, 11212, 11819, 12442, 13081, 13736, 14407, 15094, 15797

Offset: 0

Omar E. Pol, Sep 05 2011

Sequence found by reading the line from 1, in the direction 1, 16,..., in the square spiral whose vertices are the triangular numbers A000217.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A051870, A069129, A139098, A194431.

[8*n^2 +7*n + 1: n in [0..50]]; // Vincenzo Librandi, Sep 07 2011

A194268:=n->8*n^2+7*n+1: seq(A194268(n), n=0..50); # Wesley Ivan Hurt, Jul 15 2014

Table[8n^2+7n+1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{1,16,47},50] (* Harvey P. Dale, Apr 06 2014 *)

a(n)=8*n^2+7*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(0)=1, a(1)=16, a(2)=47, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Apr 06 2014
From Elmo R. Oliveira, Oct 22 2024: (Start)
G.f.: (1 + 13*x + 2*x^2)/(1 - x)^3.
E.g.f.: (1 + 15*x + 8*x^2)*exp(x). (End)

A194431 a(n) = 8n^2 - 6n - 1.

Original entry on oeis.org

1, 19, 53, 103, 169, 251, 349, 463, 593, 739, 901, 1079, 1273, 1483, 1709, 1951, 2209, 2483, 2773, 3079, 3401, 3739, 4093, 4463, 4849, 5251, 5669, 6103, 6553, 7019, 7501, 7999, 8513, 9043, 9589, 10151, 10729, 11323, 11933, 12559, 13201, 13859, 14533, 15223, 15929

Offset: 1

Omar E. Pol, Sep 05 2011

Sequence found by reading the line from 1, in the direction 1, 19, ..., in the square spiral whose vertices are the triangular numbers A000217.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A051870, A069129, A139098, A194268.

[8*n^2 - 6*n - 1: n in [1..50]]; // Vincenzo Librandi, Sep 07 2011

Table[8n^2-6n-1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,19,53},50] (* Harvey P. Dale, May 29 2021 *)

a(n)=8*n^2-6*n-1 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(-1 - 16*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 06 2011
From Elmo R. Oliveira, Jun 04 2025: (Start)
E.g.f.: 1 + (-1 + 2*x + 8*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

cp's OEIS Frontend

A139277 a(n) = n*(8*n+5).

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A195314 Centered 28-gonal numbers.

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A195315 Centered 32-gonal numbers.

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A195316 Centered 36-gonal numbers.

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A195318 Centered 44-gonal numbers.

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A195605 a(n) = (4*n*(n+2)+(-1)^n+1)/2 + 1.

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A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2*T(m) = 3*T(k) for some k.

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A139277 a(n) = n(8n+5).

A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

A194268 a(n) = 8n^2 + 7n + 1.

A194431 a(n) = 8n^2 - 6n - 1.